Calculations and Conversions

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Transcript Calculations and Conversions

Calculations and Conversions

LAT Chapter 14

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Chapter 14

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Calculations & Conversions

Chapter 14

• • Experimental methodology requires calculation.

 water consumption, record and convert temperatures, determine the floor area of a cage, detergent solution Pharmacology: the science of drugs, their composition, use, and administration • • The amount of a drug administered to a patient, based on the species and body weight (usually noted in milligrams of drug per kilogram body weight or mg/kg) = dosage.

The dosage is also influenced by an animal’s age, health, and state of excitement.

Conversions

Chapter 14

• To determine dosages, it is often necessary to convert weights and volumes from one measuring system to the other (English to metric).  For example, a technician might be required to convert kilograms to pounds, centimeters to inches, or ounces to milliliters. • You can use tables or a conversion factor by which you either multiply or divide. • Generally we use ratios to convert metric to English systems or to change units of measurement within a system.

Dose Calculations 1

Chapter 14

2.5% = 2.5 parts/100 parts, or in milligrams, 2.5 mg/100 mg 25 mg = 1000 mg 1 ml ? Ml 25 mg x ? ml = 1000 mg x 1 ml 25 mg x ? ml = 1000 mg x 1 ml 25 mg 25 mg ? ml = 1000 ml 25 ? ml = 40 ml 40 ml of dilutant must be added to prepare a 2.5% solution from 1000 mg of anesthetic.

Dose Calculations 1 (part 2)

Chapter 14

1000 mg = 300 mg 40 ml ? ml 1000 mg x ? ml = 300 mg x 40 ml 1000 mg x ? ml = 300 mg x 40 ml 1000 mg 1000 mg ? ml = 300 x 40 ml 1000 ? ml = 12 ml 12 ml of a 2.5% anesthetic solution contains 300 mg of anesthetic.

Dose Calculations 2

Chapter 14

4/100 = 4% = 40 mg/ml 40 mg = 1000 mg 1 ml ? Ml 40 mg x ? ml = 1000 mg x 1 ml 40 mg x ? ml = 1000 mg x 1 ml 40 mg 40 mg ? ml = 1000 ml 40 ? ml = 25 ml 25 ml of dilutant must be added to prepare a 4% solution from 1000 mg of anesthetic.

Dose Calculations 2 (part 2)

Chapter 14

18kg x 2.2 lb = 39.6 lb kg 1 ml = ? Ml 5 lb 39.6 lb ? ml x 5 lb = 1 ml x 39.6 lb ? ml x 5 lb = 1 ml x 39.6 lb 5 lb 5 lb ? ml = 39.6 ml 5 ? ml = 7.92 ml

Conversion of % Solutions

Chapter 14

• • • • • • Solutions are homogeneous mixtures of two or more constituents. A solution can take the form of a gas, a liquid, or a solid.

Gaseous solutions are made by dissolving one gas in another. Since all gases mix in all proportions, any mixture of two or more gases is homogeneous and is a solution.

In solid solutions one component is randomly dispersed on an atomic or molecular scale throughout another component.

Liquid solutions are made by dissolving a gas, liquid, or solid in a liquid.  If the liquid is water, the solution is called an aqueous solution.

Solution Components

Chapter 14

• • • • In making up a solution, at least one constituent must be a solute and one a solvent. Substance present in larger amount = solvent.  Substance present in the smaller amount = solute.

% of solute may = % by weight or % by volume. % by weight = total solution mass of solute.  % weight in volume (w/v) expresses the no. of grams of solute in 100 milliliters (ml) of solvent.  % weight in weight (w/w) expresses the no. of grams of solute in 100 grams of solvent.

Solution Components pt.2

Chapter 14

• • 5 g of dextrose dissolved in sufficient water to bring the final volume to 100 ml is a 5 % dextrose solution by weight in volume (w/v). Set up in ratio fashion to translate them into a more useful expression of mg/g or mg/ml.

A simple way to translate % 5 g = ? g into mg/ml or mg/g is to move 100 ml 1 ml the decimal point one place to the right. 100 ml x ? g = 1 ml x 5 g For example: ? = 0.05 g/ml 5.0 % = 50 mg/ml or, since there are 1000 mg/g 2.2 % = 22 mg/ml ? = 50 mg/ml 20.0 % = 200 mg/ml

Percent by Volume

Chapter 14

• % by volume is the percent of the final solution volume represented by the volume of the solute taken to make the solution (v/v). • For example, a 12 % alcohol solution by volume would be a solution made from 12 ml of alcohol and enough solvent to bring the total volume up to 100 ml.

Molar Solutions

Chapter 14

• Molecular weight = sum of atomic weights of atoms in a molecule.  A 1-molar solution contains the molecular weight (MW) in grams of that compound per liter of solution. • To make an aqueous solution of cobalt chloride (CoCl 2 ) it would be necessary to calculate the MW of CoCl 2 Atomic mass of Co = 58.85 x 1 = 58.85

from the Periodic Table: Atomic mass of Cl = 35.5 x 2 = 71.00

Molecular Weight in g of CoCl 2 = 129.85

 1 M solution of CoCl 2 , 129.85 g of CoCl 2 are weighed & dissolved in water to a total volume of 1000 ml, or 1 L.  If 2 moles are needed, 129.85 x 2, which equals 259.7 g/L. This weight of cobalt chloride is added to a 1-liter volumetric flask, & the flask is filled to the one liter mark with water, giving a final volume of 1 liter of a 2M solution of CoCl 2 .

Molality

Chapter 14

• • • No. of moles of solute per 1000 grams of solvent. Therefore, a 1-molar aqueous solution of sucrose, C 12 H 22 O ll , contains 342.2 g of sucrose in 1000 g of water. This is calculated by multiplying the atomic mass of each substance by its respective number of moles.

 C 12 H 22 O 11 Atomic Mass of C = 12.0 x 12 = 144.0

Atomic Mass of H = 1.01 x 22 = 22.2

Atomic Mass of O = 16.0 x 11 = 176.0

Molecular weight of sucrose in g = 342.2

Statistics, Calculations and Conversions

Chapter 14

Additional Material and Review

Statistics, Calculations and Conversions

Chapter 14

Statistics is a branch of mathematics that deals with the collection, analysis, interpretation, and presentation of numerical data.

The collection of the raw data is the most important step in the use of statistics to make valid conclusions about the topic being studied.

Statistics are usually used to determine if any differences exist between treated groups and control groups of animals.

Statistics, Calculations and Conversions

Chapter 14

Terminology

Validity

- conclusions are correctly drawn from the data

Random, Random Selection, Random Number Tables

- members in a given set have an equal probability of an event occurring

Populations

- sets of animals or objects that have some common characteristic

Samples

- subsets of the population

Parameters

- measurable characteristics

Bias

- tendency to deviate from the truth

Statistics, Calculations and Conversions

Chapter 14

Conversions - change weights or volumes from one measuring system to another using a conversion factor 1 inch = 2.54 cm 1 gallon = 3.87 liters 1 kilogram = 2.2 pounds 1 mile = 1.6 kilometer

Statistics, Calculations and Conversions

Chapter 14

Dose calculations: Reconstitute 1/2 gram antibiotic powder to give a 5% solution and give a 25 mg dose to the animal 1/2 gram = 500 mg 1 ml of H 2 O = 1 gram 5% = 5 parts/100 parts or 5 mg/100 mg or 50 mg/1000 mg = 50mg/ml 50 mg = 500 mg 1 ml ? ml 50 mg(? ml) = 500 mg (1ml) 50 mg (?ml) = 10 50 mg = 500 mg (1 ml) 50 mg ? ml = 500 mg 50 mg ? ml

Statistics, Calculations and Conversions

Chapter 14

Dose calculations: Reconstitute 1/2 gram antibiotic powder to give a 5% solution and give a 25 mg dose to the animal 500 mg 10 ml 500 mg (? ml) ? ml ? ml = = = = 25 mg ? ml 25 mg (10 ml) 25 mg (10 ml) 500 0.5

Statistics, Calculations and Conversions

Chapter 14

Solutions are homogenous mixtures of 2 or more constituents and can be gas, liquid, or solid.

Liquid solutions are made by dissolving gas, liquid, or solid into a liquid; if water is used, this is referred to as an aqueous solution.

Statistics, Calculations and Conversions

Chapter 14

Solutions = solute and solvent Percent by weight = total mass contributed by solute Percent weight in volume (w/v) = grams of solute/100 mls solvent Percent weight in weight (w/w) = grams of solute/100 grams solvent Percent by volume (v/v) = percent of final solution volume made up by the solute volume

Statistics, Calculations and Conversions

Chapter 14

Mole = quantity of atoms = 6.022 X 10 23 = Avagadro’s number A one molar solution = 1 M = molecular wt in grams/liter solution Molecular weights are listed in the Periodic Table in the upper right CCl 4 1 Carbon + 4 Chloride = 12g X 1 mole + 35.5 X 4 = 12 + 142 = 154 grams

Statistics, Calculations and Conversions

Chapter 14

Normality = N = the number of equivalents of solute/liter of solution The number of equivalents = actual weight of the substance/equivalent weight Equivalent weight = Atomic weight/valence

Statistics, Calculations and Conversions

Chapter 14

Valence (Net positive valence) = (Valence of + element) X (its subscript) NaCl K 2 CO 3 CaSO 4 AlCl 3 Fe 2 (SO 4 ) 3 1 X 1 = 1 1 X 2 = 2 2 X 1 = 2 3 X 1 = 3 3 X 2 = 6 Equivalent wt = Molecular weight/Net positive valence 58.5/1 138/2 136/2 133.5/3 400/6

58.5/1

Statistics, Calculations and Conversions

Chapter 14

138/2 136/2 133.5/3 400/6 58.5

69 68 44.5

66.6

The number of equivalents = actual weight of the substance/equivalent weight.

Normality = N = the number of equivalents of solute/liter of solution.

Normality = molarity X net positive valence.

Therefore, normality will always be equal to or greater than molarity.

Chapter 14

THE END